[SOLVED] Equation of a plane - Given z intercept and parametric equation

Sep 2009
83
0
Hey,

I have been working on this question for the past couple days (searching the internet, searching for examples in textbook etc.) but I cannot figure out how to answer this question. I will likely have follow up questions to your solution, so please keep an eye on this thread following helping me!

Question
Determine a scalar equation of a plane that contains the line r = (3, -1, 1) + t(2, 4, -5) and whose z-intercept is -2.

Solution
The textbook gives a hint in the back of the book that I need 2 direction vectors in order to find the solution. The question is giving me one direction vector "(2, 4, -5)" but I am unsure of how to find the second one. After I find the second direction vector (which I am unsure how to do) I find the cross product of the two direction vectors to find the normal. After having the normal I think I need a point on the plane The z intercept is listed, so I know a point is (0, 0, -2), right?
Knowing this I can simply sub into the D = -Ax - By - Cz. (which A B C and are from the normal and the x y z are from the point on the plane)
This is how I am understanding what I need to do. PLEASE correct me if I am mistaken here! I am unsure how to do certain steps in my explanation, which is what I am needing help with.

Thank you for helping!!!
 
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earboth

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Hey,

I have been working on this question for the past couple days (searching the internet, searching for examples in textbook etc.) but I cannot figure out how to answer this question. I will likely have follow up questions to your solution, so please keep an eye on this thread following helping me!

Question
Determine a scalar equation of a plane that contains the line r = (3, -1, 1) + t(2, 4, -5) and whose z-intercept is -2.

Solution
The textbook gives a hint in the back of the book that I need 2 direction vectors in order to find the solution. The question is giving me one direction vector "(2, 4, -5)" but I am unsure of how to find the second one. After I find the second direction vector (which I am unsure how to do) I find the cross product of the two direction vectors to find the normal. After having the normal I think I need a point on the plane The z intercept is listed, so I know a point is (0, 0, -2), right?
Knowing this I can simply sub into the D = -Ax - By - Cz. (which A B C and are from the normal and the x y z are from the point on the plane)
This is how I am understanding what I need to do. PLEASE correct me if I am mistaken here! I am unsure how to do certain steps in my explanation, which is what I am needing help with.

Thank you for helping!!!
1. The line passes through P(3, -1, 1) and the plane passes through Q(0,0,-2). Then the vector \(\displaystyle \overrightarrow{PQ}\) is the second direction vector which span he plane.

2. \(\displaystyle p: \vec r = (3,-1,1) + t(2,4,-5)+s(-3,1,-3)\)

3. Now calculate the cross product of the two direction vectors:

\(\displaystyle (2,4,-5) \times (-3,1,-3) = (-7,21,14) = 7(-1,3,2)\)

4. Use the "reduced" vector to determine the equation of the plane.
 
Sep 2009
83
0
1. The line passes through P(3, -1, 1) and the plane passes through Q(0,0,-2). Then the vector \(\displaystyle \overrightarrow{PQ}\) is the second direction vector which span he plane.

2. \(\displaystyle p: \vec r = (3,-1,1) + t(2,4,-5)+s(-3,1,-3)\)

3. Now calculate the cross product of the two direction vectors:

\(\displaystyle (2,4,-5) \times (-3,1,-3) = (-7,21,14) = 7(-1,3,2)\)

4. Use the "reduced" vector to determine the equation of the plane.
Thank you! I just figured this out on my own, but thank you for answering so quickly!